Question

Hello,

What is the general solution of the following differential equation?

(1 + (x^3/y)sin^2x)dx + (1/y)(x + (1/cos^2(2y))dy = 0;
I used the integrating factor, but I'm stuck at the following step:
u(x) [-x^3sin^2x + 1] = u'(x) [x + (1/cos^2(2y)]

Thanks in advance!

Answer 1

can you rewrite it using the equation editor on the bottom? It's hard to see what you have there, but it looks like an exact differential equation.

Answer 2

yes, im trying to make it into one.

\[u(x) [-x ^{3}\sin ^{2} x + 1] = u'(x) [x + (1/\cos ^{2}(2y)]\]

Answer 3

No, I mean the original differential equation.

Answer 4

The original is a non-exact equation, but I'm trying to use the integrating factor method to make it into an exact equation.

Answer 5

Any idea on how to simplify what i have to solve for u(x)?

Answer 6

Did you use substitution method to put it into a first order differential equation? I don't quite understand what you're doing. Sorry. lol. That's why I'm having troubles helping you right now.

Answer 7

First, I checked if the equation was an exact one. I found out that it wasn't, so I multiplied both sides of the equation by u(x) = integrating factor. This gave me a new equation which lets us know that it is in fact an exact equation.

Answer 8

Now i just have to solve for the integrating factor.

Answer 9

The original equation is:
\[(1 + (x^3/y)\sin^2x)dx + (1/y)(x + (1/\cos^2(2y))dy = 0\]

Answer 10

\[
\left(1+{x^3\over y}\sin^2x\right)+{1\over y}\left(x+{1\over\cos^2(2y)}\right){dy\over dx}=0
\]
is this what it is!!!!

Answer 11

can read it properly...

Answer 12

what?

Answer 13

now i can nvermind.

Answer 14

it was just showing a bunch of symbols before.

Answer 15

ok now?

Answer 16

yes.

Answer 17

so, what was your step for this?

Answer 18

do you have any idea how to go on from where im stuck? or any other suggestions?

Answer 19

u(x)[−x 3 sin 2 x+1]=u ′ (x)[x+(1/cos 2 (2y)]

Answer 20

could you write that clearly like I did.

Answer 21

i did before, let me try again

Answer 22

\[u(x) [-x ^{3}\sin ^{2} x + 1] = u'(x) \left[x +{1\over\cos ^{2}(2y)}\right]\]

Answer 23

yes that is it.

Answer 24

im trying to solve for the integrating factor u(x).

Answer 25

problem 1) in the first paranthesis... something smells bad

Answer 26

how would you solve it?

Answer 27

\[M(x,y)=1+{x^3\over y}\sin^2x\qquad N(x,y)={x\over y}+{1\over y\cos^2(2y)}\\
M_y=-{x^3\over y^2}\sin^2x\qquad\qquad\quad N_x={1\over y}\]

Answer 28

I got those values, therefore indicating that the equation is not exact.

Answer 29

\[
N_x-M_y={1\over y}\left(1+{x^3\over y}\sin^2x\right)={M\over y}\\
\therefore \ln[u(y)]=\int{1\over y}dy=\ln y\\
\boxed{u(y) = y}
\]

Answer 30

The next part is very tedious especially to put on this IM area.

Answer 31

why did u subtract?

Answer 32

well thank you and im going to absorb this as soon as possible!

Answer 33

then we create our new DE
\[
(y+x^3\sin^2x)+\left[x+{1\over\cos^2(2y)}\right]{dy\over dx}=0
\]
let us check if this is exact
\[M(x,y)=y+x^3\sin^2x\qquad N(x,y)=x+{1\over\cos^2(2y)}\\
M_y=1\qquad\qquad\qquad N_x=1\qquad{\rm=exact}\\
F(x,y)=\int M\,dx+g(y)\qquad F(x,y)=\int N\,dy+g(x)
\]

Answer 34

1)\[F=\int\left[x+\sec^2(2y)\right]dy+g(x)\\
\boxed{\large F(x,y)=xy+{\tan(2y)\over2}+g(x)}
\]

Answer 35

but \(F_x=M\)
\[
y+g'(x)=y+x^3\sin^2x\\\implies
g(x)=\int(x^2\sin^2x)dx
\]

Answer 36

correction, make it a cube

Answer 37

ok

Answer 38

this is the only worse part..
find g(x) and complete your F(x,y)

Answer 39

i couldnt get that far.

Answer 40

ill try to get there though; thanks!

Answer 41

did you follow the steps?

Answer 42

yw

Answer 43

yup, i had a prob with understanding why subtracted the partial derivatives earlier.

Answer 44

you*

Answer 45

to find the missing factor u(y) to make the equation exact.

Answer 46

i get that but y subtract them.erhaps im missing something.

Answer 47

we have to see what difference would make it a factor of one of these N(x,y) or M(x,y)

Answer 48

subtraction gives you the missing portion :)

Answer 49

i really appreciate it! thanks a bunch!

Answer 50

welcome another bunch

Answer 51

how do u give a badge?

Answer 52

you did! when you clicked the "best response"

Answer 53

haha ok

Answer 54

thanks

Answer 55

peace out

Answer 56

np peace!